Question

Evaluate \[ \int \frac{3x + 2}{(x^2 + 1)(x - 2)} \, dx \]

Hint:

When using partial fractions, break the integrand into simpler terms to make integration easier.

Answer 1

Step 1: Decomposing the integrand.
We use partial fraction decomposition to express \( \frac{3x + 2}{(x^2 + 1)(x - 2)} \) as the sum of simpler fractions. Assume: \[ \frac{3x + 2}{(x^2 + 1)(x - 2)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x - 2} \] Step 2: Solving for the constants.
Multiply both sides by \( (x^2 + 1)(x - 2) \) to eliminate the denominators: \[ 3x + 2 = (Ax + B)(x - 2) + C(x^2 + 1) \] Expanding both sides: \[ 3x + 2 = (Ax^2 - 2Ax + Bx - 2B) + Cx^2 + C \] \[ 3x + 2 = (A + C)x^2 + (-2A + B)x + (-2B + C) \] Step 3: Equating coefficients.
Compare the coefficients of \( x^2 \), \( x \), and the constant term on both sides:
\( A + C = 0 \)

\( -2A + B = 3 \)

\( -2B + C = 2 \)

Step 4: Solving the system of equations.
From \( A + C = 0 \), we get \( C = -A \).
Substitute \( C = -A \) into the second and third equations: \[ -2A + B = 3 \quad \text{and} \quad -2B - A = 2 \] Solve the system:
\( B = 3 + 2A \)

Substitute into the second equation: \[ -2(3 + 2A) - A = 2 \] \[ -6 - 4A - A = 2 \quad \Rightarrow \quad -5A = 8 \quad \Rightarrow \quad A = -\frac{8}{5} \]

Substitute \( A = -\frac{8}{5} \) into \( C = -A \): \[ C = \frac{8}{5} \]

Substitute \( A = -\frac{8}{5} \) into \( B = 3 + 2A \): \[ B = 3 + 2\left( -\frac{8}{5} \right) = 3 - \frac{16}{5} = \frac{-1}{5} \]
Step 5: Substituting back into the integral.
Now substitute the values of \( A \), \( B \), and \( C \) into the partial fraction decomposition: \[ \frac{3x + 2}{(x^2 + 1)(x - 2)} = \frac{-\frac{8}{5}x - \frac{1}{5}}{x^2 + 1} + \frac{\frac{8}{5}}{x - 2} \] Step 6: Integrating each term.
Now integrate each term: \[ \int \frac{-\frac{8}{5}x - \frac{1}{5}}{x^2 + 1} \, dx + \int \frac{\frac{8}{5}}{x - 2} \, dx \] Using the standard integrals: \[ \int \frac{x}{x^2 + 1} \, dx = \frac{1}{2} \ln(x^2 + 1) \] and \[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x) \] We obtain: \[ \frac{-8}{5} \left( \frac{1}{2} \ln(x^2 + 1) \right) - \frac{1}{5} \tan^{-1}(x) + \frac{8}{5} \ln|x - 2| + C \]